One of the nicest theorems in linear algebra is the one that a matrix satisfies its own characteristic polynomial, the so-called Cayley-Hamilton theorem.

What is a good way to prove it? In particular, does this elegant proof go through.

I am hopeful that it is quite trivial. Namely, since the characteristic polynomial is $\rm{det}(A-\lambda I)$, if we plug in $A$ for $\lambda$, we of course get $\rm{det}0=0$.

If so, this seems like one of the easiest times a couple of mathematicians got away with a major theorem.

To be precise, is there any problem with replacing $\lambda$, which usually denotes a scalar, with the matrix in question $A$.

  • $\begingroup$ This should answer your question: math.stackexchange.com/questions/1141648/… $\endgroup$
    – teddy
    May 29, 2020 at 5:06
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    $\begingroup$ As you study mathematics, you will see a number of theorems that in retrospect are trivial, often due to notation we've developed which makes the theorem easier to prove. Cayley-Hamilton is not such a theorem. $\endgroup$ May 29, 2020 at 5:34
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    $\begingroup$ One day, I also thought about replacing $\lambda$ by $A$ to prove the theorem. But then it occurred to me that det$(A-\lambda I) $ is a characteristic polynomial where $\lambda$ is scaler. How can I put $A$ a matrix ( not one by one) in place of a scaler? The theorem can be proven easily in case A is diagonalizable. $\endgroup$
    – Koro
    May 29, 2020 at 6:09
  • $\begingroup$ There is no such thing as the proof of the Cayley-Hamilton theorem; indeed it is the algebraic theorem for which I think I have seen the widest variety of proofs. Other theorems with many different proofs that come to mind are Pythagoras' theorem and quadratic reciprocity. But whereas Pythagoras' theorem becomes almost trivial in certain settings (like in the context of Euclidean vector spaces) I don't think this is the case with Cayley-Hamilton; indeed a tradeoff between generality, technical simplicity and conceptual elementariness is always involved. Different proofs serve different needs $\endgroup$ May 4, 2021 at 9:51
  • $\begingroup$ Thanks @MarcvanLeeuwen I have edited slightly to reflect it. $\endgroup$
    – user403337
    May 8, 2021 at 3:44

3 Answers 3


"The" proof of the Cayley-Hamilton Theorem involves invariant subspaces, or subspaces that are mapped onto themselves by a linear operator. If $T$ is a linear operator on a vector space $V$, then a subspace $W\subseteq V$ is called a $T$-invariant subspace of $V$ if $T(W)\subseteq W$, i.e. if $T(v)\in W$ for every $v\in W$. Some examples of $T$-invariant subspaces you might be familiar with are $\{0\}, N(T), R(T), V$, and $E_\lambda$ for any eigenvalue $\lambda$ of $T$. For a linear operator $T$ and any nonzero $x\in V$, then the subspace $$ W=\textrm{span}(\{x,T(x),T^2(x),\dots\})$$ is called the $T$ cyclic subspace of $V$ generated by $x$, and one can show that $W$ is the smallest $T$-invariant subspace containing $x$. Cyclic subspaces can be used to establish the Cayley-Hamilton Theorem. In fact, the existence of a $T$-invariant subspace allows us to define a new linear operator whose domain is this subspace, i.e. the restriction $T_W$ of $T$ to $W$ is a linear operator from $W$ to $W$. These two operators are linked in the sense that the characteristic polynomial of $T_W$ divides the characteristic polynomial of $T$. You can show this by choosing your favorite ordered basis for $W$ and extending it to an ordered basis for $V$, then taking the matrix representations of $T$ and $T_W$, and computing the characteristic polynomial of $T$, one will see that the characteristic polynomial of $T_W$ can be recovered.

The last tool we will need is how to gain information about the characteristic polynomial of $T$ from the characteristic polynomial of $T_W$. Cyclic subspaces are useful in this sense because the characteristic polynomial of the restriction of a linear operator $T$ to a cyclic subspace can be computed. In fact, if $T$ is a linear operator on a finite-dimensional vector space $V$, then if $W$ is the $T$ cyclic subspace of $V$ generated by a nonzero $v\in V$, and letting $k=\textrm{dim}(W)$, then we have that:

  1. $\{v,T(v),T^2(v),\dots,T^{k-1}(v)\}$ is a basis for $W$
  2. If $a_0v+a_1T(v)+\cdots+a_{k-1}T^{k-1}(v)+T^k(v)=0$, then the characteristic polynomial of $T_W$ is $f(t)=(-1)^k(a_0+a_1t+\cdots+a_{k-1}t^{k-1}+t^k)$

I will omit the proof for the above theorem unless requested, since the main goal is the proof of the Cayley-Hamilton Theorem, which states that:

Let $T$ be a linear operator on a finite-dimensional vector space $V$, and let $f(t)$ be the characteristic polynomial of $T$. Then $f(T)=T_0$, the zero transformation. That is, $T$, "satisfies" its characteristic equation.

Proof: To show that $f(T)(v)=0$ for all $v\in V$. If $v=0$, we are done since $f(T)$ is linear, so suppose $v\neq 0$, and let $W$ be the $T$-cyclic subspace generated by $v$ with dimension $k$. By the theorem above, there exist scalars $a_0,\dots,a_{k-1}$ such that $$a_0v+a_1T(v)+\cdots+a_{k-1}T^{k-1}(v)+T^k(v)=0 $$ and the characteristic polynomial for $T_W$ is: $$ g(t)=(-1)^k(a_0+a_1t+\cdots+a_{k-1}t^{k-1}+t^k)$$ Combining these two inequalities yields: $$g(T)(v)=(-1)^k(a_0I+a_1T+\cdots+a_{k-1}T^{k-1}+T^k)(v)=0 $$ We know that this polynomial divides the characteristic polynomial of $T$, $f(t)$, thus there exists a polynomial $q(t)$ such that $f(t)=q(t)g(t)$, so: $$ f(T)(v)=q(T)g(T)(v)=q(T)(g(T)(v))=q(T)(0)=0$$ The Cayley-Hamilton Theorem for Matrices is then a corollary to the Cayley-Hamilton Theorem stated above.

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    $\begingroup$ It was, in retrospect, wishful thinking. It's quite subtle after all. Impressive @teddy $\endgroup$
    – user403337
    May 29, 2020 at 5:44
  • $\begingroup$ Thanks, I provided a link in the comments as to why you can't just plug $A$ for $\lambda$ so I hope that helps you as well @ChrisCuster $\endgroup$
    – teddy
    May 29, 2020 at 5:46
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    $\begingroup$ Yeah I saw that. The expression you get is basically meaningless. I made the same mistake as the OP on the link. It was tempting. But alas. I hadn't revisited this in a while, and apparently never resolved it in a way that I could remember. Speaking of which, my advisor once referred to "the" proof as being one that you can remember. Sometimes it hasn't come along yet. Thanks. $\endgroup$
    – user403337
    May 29, 2020 at 6:03

In older textbooks, the usual proof is to substitute $A$ into the characteristic polynomial $p(x)=\det(xI-A)$ in a correct manner. The proof is basically a one-liner: since $\operatorname{adj}(Ix-A)(Ix-A)=p(x)I$, we have $p(A)=0$ by the factor theorem. It probably offers the best explanation why the theorem holds (it's because substitution works if you do it correctly), but it involves many subtle points that beginners in linear algebra may find difficult to understand.

Among all linear algebra textbooks that I have ever read, Mac Duffee's Vectors and Matrices offers the clearest explanation to the above proof (see chapter IV). The following resources are also useful:


The cleanest proof uses the adjugate, an algebraic complement of the matrix A.

See it in https://en.wikipedia.org/wiki/Cayley–Hamilton_theorem under "A direct algebraic proof".

Recall the definition of the adjugate. Replace every matrix element $a_{ij}$ of the $n\times n$ matrix $A$ with the determinant of the matrix $A$ with the row $i$ and column $j$ skipped, and a chessboard sign. Then transpose the matrix, to obtain the adjugate $\text{adj}(A)$. Then $A \ \text{adj}(A) = \det A I$. This is immediate from the expansion of $\det A$ by a row or a column.

Let the characteristic polynomial be $p(t) = \det(tI-A)= \sum c_i t^i I$ where $c_n=1$.

Take now $B = \text{adj}(tI - A)$ so $(tI - A)B = p(t) I$.

Expand $B$ in powers of $t$, as $B = \sum_{i=0}^{n-1} t^i B_i$, to get $B_{i-1}-AB_i = c_i I$ where $B_{n}=B_{-1}=0$.

Multiply from the left by $A^i$ to get $c_i A^i = A^i B_{i-1}-A^{i+1}B_i$.

Sum, to get after a telescopic cancellation $p(A)=\sum c_i A^i = 0$.

The nice part of this proof is that it gives an explicit expression for each term $c_i A^i$ using the determinant expressions for $B_i$, so it is amenable to generalizations, for instance to non commutative coefficients.

My high school linear algebra book, in Eastern Europe, was 1/4 in thick, printed on rough paper, but had this proof in it. Most of today's 2in thick undergrad books are very colorful, but skip the proof, as too hard.


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